For what value of a the quadratic ax^2 +(2a-1)x +a+4 is always positive

Solution:  We have  ,  $ax^{2}+(2a-1)x+(a+4)$

$\\ \\ \Rightarrow \hspace{1cm}D> 0\hspace{0.5cm} and \hspace{0.5cm}a> 0\\ \\ \Rightarrow \hspace{1cm}D=(2a-1)^{2}-4a(a+4)< 0\\ \\ \Rightarrow \hspace{1cm}-8a+1< 0\Rightarrow a> \frac{1}{8}\\ \\ \Rightarrow \hspace{1cm}a\in (\frac{1}{8},\infty)$

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